The entropy $S$ of a system is defined as $$S = k\ln \Omega.$$ What precisely is $\Omega$? It refers to "the number of microstates" of the system, but is this the number of all accessible microstates or just the number of microstates corresponding to the systems current macrostate? Or is it something else that eludes me?

  • 1
    $\begingroup$ Uh, ambiguity is the definition of entropy. $\endgroup$
    – Hot Licks
    Commented Apr 4, 2019 at 12:06

3 Answers 3


Entropy is a property of a macrostate, not a system. So $\Omega$ is the number of microstates that correspond to the macrostate in question.

Putting aside quantization, it might appear that there are an infinite number of microstates, and thus the entropy is infinite, but for any level of resolution, the number is finite. And changing the level of resolution simply multiplies the number of microstates by a constant amount. Since it is almost always the change in entropy, not the absolute entropy, that is considered, and we're taking the log of $\Omega$, it actually doesn't matter if the definition of S is ambiguous up to a constant multiplicative factor, as that will cancel out when we take dS. So with a little hand waving (aka "normalization"), we can ignore the apparent infinity of entropy.

  • $\begingroup$ +1; You may add for completeness that $\Omega$ is actually the volume occupied by all the possible microstates corresponding to the given macrostate, in phase space. I Always found this expression clearer, since both positions and momentum are continuous variables, the number of microstates available would be infinite. $\endgroup$ Commented Apr 3, 2019 at 14:59
  • 1
    $\begingroup$ @Runlikehell I was trying to get at the infinity problem with my last paragraph, but I think I'll make that a bit more clear. $\endgroup$ Commented Apr 3, 2019 at 15:31
  • $\begingroup$ +1. "Entropy is a property of a macrostate, not a system." is completely clear, but I have never heard that before. It clarified my thinking a lot. $\endgroup$
    – M. Winter
    Commented Apr 4, 2019 at 8:19
  • $\begingroup$ @M. Winter, in physics, the macrostate " in question " is the state in which system exists or more accurately spend most of the time, so we (or the system) must choose the macrostate with the highest stat. weight, that is, highest $ \Omega$. Analogy: the extremum of the function (entropy) is a property of the function (system) and not a particular function value of particular argument value $\endgroup$ Commented Apr 4, 2019 at 10:32

Entropy logarithmically measure of the number of microscopic states corresponding to some specific macroscopically-observable state, not the system as a whole. Put another way: systems that have not yet found their equilibrium state, when left alone, increase their entropy. This would not be possible if the system had the same entropy for all macrostates.

Indeed, the driving principle of entropy in modern stat-mech says that we have some uncertainty about the underlying microscopic state of the system and that from a certain perspective (basically, the one where every macroscopic quantity we can determine is conserved) we can treat nature as simply choosing a microstate uniformly at random. (We have to tread carefully about what exactly uniformly means here but an “obvious” choice seems to replicate certain nice features, like that metals will have specific heats that look like $3R$ where $R$ is the gas constant—a result that I want to say is due to Einstein but I am not 100% sure.)

As a result of this principle of nature picking microstates at random, our equilibrium state is the macrostate which contains the most microstates, and our regression to equilibrium is a process of macrostates getting larger and larger.


Entropy is a matter of perspective.

You pick a way to describe a system at large scales. This effectively subdivides the system into macrostates, or "macroscopic states".

Each of these macroscopic states corresponds to a number of "microstates"; different configurations of the system that are clumped together in one macrostate.

If, for each macrostate, you take the log of the number of microstates in it, the principle of Entropy is that whatever macrostate it is in, it will move towards macrostates with a higher value almost certainly.

Now you can move to a lower Entropy value only by increasing the Entropy of another system. This basically consists of merging the two systems into one and applying the first rule.

The number of microstates multiply when they are combined; if we have two systems A and B, and they have macrostates A_0 and B_0 with 7 and 10 microstates apiece, the system A+B with macrostate A_0+B_0 has 70 microstates (7*10).

Taking the log of the number of microstates simply allows us to use addition instead of multiplication; the entropy of $\log(7)$ and $\log(10)$ add to $\log(7)+\log(10)$ = $\log(7*10)$.

Any function that has the property that $f(a*b)=f(a)+f(b)$ will do just as well, which is why we don't care what the base of our logarithm is.

The fun part is that this applies regardless of how you clump the microstates into macrostates so long as you do the clumping before the experiment. So we go and pick sensible macrostates that correspond to things we care about, and the result holds. Crazy choice of macrostates don't actually help us; the vast majority of the possible configuration space of any system is completely useless chaos, only a ridiculously small fraction of the system configuration space is going to be "useful", and no matter how we label it that space is going to have very few microstates in it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.